Chapter 1.6, Problem 33PPS

a.

To determine

To find the maximum and minimum circumference

Answer to Problem 33PPS

Minimum circumference is 25 inches and maximum circumference is 31 inches

Explanation of Solution

Given information:

  Diameter is 8 to 10 inches

Calculation:

For minimum circumference,

  Circumference= $2\pi r$

Solving,

  Circumference= $2\pi (\frac{8}{2})$   ( D = 2r )

So, circumference= $8\pi$

Substituting value,

  Circumference= $8(3.14)$

Minimum circumference is 25.12.

For maximum circumference,

  Circumference= $2\pi r$

Solving,

  Circumference= $2\pi (\frac{10}{2})$   ( D = 2r )

So, circumference= $10\pi$

Substituting value,

  Circumference= $10(3.14)$

Maximum circumference is 31.4.

Hence minimum circumference is 25 inches and maximum circumference is 31 inches.

b.

To determine

To find the range of possible areas

Answer to Problem 33PPS

Range of area is 50.24 to 78.5 inches²

Explanation of Solution

Given information:

  Diameter is 8 to 10 inches

Calculation:

Radius= $\frac{8}{2}$ to $\frac{10}{2}$

So,

  radius is 4 inches to 5 inches

Area for minimum,

  Area= $\pi r^2$

Substituting values,

  Area= $\pi {(4)}^2$

Solving,

  Area= $16(3.14)$

Minimum area is 50.24 inches².

Area for maximum,

  Area= $\pi r^2$

Substituting values,

  Area= $\pi {(5)}^2$

Solving,

  Area= $25(3.14)$

Maximum area is 78.5 inches².

Hence, range of area is 50.24 to 78.5 inches².